tìm X: 3 x [1/5x8 + 1/8x11 + 1/11x14 + ... + 1/97x100 + X] = 319/100
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
26 tháng 3 2018
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}.\frac{1}{3}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{3}{160}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{160}\)
\(\Rightarrow\frac{1}{x+3}=\frac{31}{160}\)
\(\Rightarrow160=31x+93\)
\(\Rightarrow31x=67\)
\(\Rightarrow x=\frac{67}{31}\)
25 tháng 5 2021
S= 1/2x5 + 1/5x8 + 1/8x11 + 1/11x14 + .... + 1/97x100
S = 1/2 x 1/5 + 1/5 x 1/8 + 1/8 x 1/11 + 1/11 x 1/14 + .......+ 1/97 x 1/100
S = 1/2 x ( 1/5 + 1/5 x 1/8 + 1/8 x 1/11 + 1/11 x 1/14 + .......+ 1/97 ) x 1/100
S = 1/2 x 1/100
S = 1/200
~ Hok T ~
HT
25 tháng 5 2021
\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{97\cdot100}\)
\(S=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+....+\frac{3}{97\cdot100}\right)\)
\(S=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\cdot\frac{49}{100}\)
\(S=\frac{49}{300}\)
7 tháng 7 2020
Ta có: \(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\frac{3}{7}=\frac{1}{21}\)
\(\Leftrightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}\cdot\frac{7}{3}=\frac{7}{63}=\frac{1}{9}\)
Vậy: \(x=\frac{1}{9}\)
BV
1
DP
27 tháng 6 2017
\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+....+\frac{1}{97\cdot100}\)
\(=\frac{5-2}{2\cdot5}+\frac{8-5}{5\cdot8}+\frac{11-8}{8\cdot11}+...+\frac{100-97}{97\cdot100}\)
\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{3}\cdot\frac{49}{100}=\frac{49}{300}\)
PH
0
\(3\times\left(\frac{1}{5\times8}+\frac{1}{8\times11}+....+\frac{1}{97\times100}+x\right)=\frac{319}{100}\)
\(\Rightarrow\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{97\times100}\right)+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{19}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow3\times x=\frac{319}{100}-\frac{19}{100}\)
\(\Rightarrow3\times x=3\)
\(\Rightarrow x=3:3\)
\(\Rightarrow x=1\)
Vậy x = 1